Fionn observed that $91\%$ of the background checks he processes are for potential employees who have already interviewed. Let $B$ be the number of background checks Fionn processes to get his first background check for a potential employee who has not yet interviewed for the job. Assume that the interview statuses are independent. Find the probability that the $8^{\text{th}}$ background check that Fionn processes is the first for which the potential employee has not yet interviewed. You may round your answer to the nearest hundredth. $P(B=8)=$
Answer: Without a fancy calculator For each background check: $P({\text{not yet interviewed}})=0.09$ $P(\text{done with interview}})=0.91$ If Fionn's $8^{\text{th}}$ background check is for the first potential employee who has not yet interviewed, then Fionn must first process $7$ background checks for potential employees who have already interviewed, then process one for someone who has not yet had an interview. $\begin{aligned} P(B=8)&=P(\text{DDDDDDD}}{\text{N}}) \\\\ &=(0.91})(0.91})\cdots(0.91})({0.09}) \\\\ &=(0.91)^7(0.09) \\\\ &\approx0.0465 \end{aligned}$ $P(B=8)\approx 0.0465\approx0.05$